Problem: You have found the following ages (in years) of all 4 lions at your local zoo: $ 1,\enspace 6,\enspace 1,\enspace 1$ What is the average age of the lions at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 4 lions at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{1 + 6 + 1 + 1}{{4}} = {2.3\text{ years old}} $ Find the squared deviations from the mean for each lion. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $1$ year $-1.3$ years $1.69$ years $^2$ $6$ years $3.7$ years $13.69$ years $^2$ $1$ year $-1.3$ years $1.69$ years $^2$ $1$ year $-1.3$ years $1.69$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{1.69} + {13.69} + {1.69} + {1.69}} {{4}} $ $ {\sigma^2} = \dfrac{{18.76}}{{4}} = {4.69\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{4.69\text{ years}^2}} = {2.2\text{ years}} $ The average lion at the zoo is 2.3 years old. There is a standard deviation of 2.2 years.